Solve Using the Quadratic Formula 4x^2+2x=0

$4 x^{2} + 2 x = 0$

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = 4$ , $b = 2$ , and $c = 0$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (4 \cdot 0)}}{2 \cdot 4}$

Step 3

Simplify.

Simplify the numerator.

Raise $2$ to the power of $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 4 \cdot 0}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot 0$ .

Multiply $- 4$ by $4$ .

$x = \frac{- 2 \pm \sqrt{4 - 16 \cdot 0}}{2 \cdot 4}$

Multiply $- 16$ by $0$ .

$x = \frac{- 2 \pm \sqrt{4 + 0}}{2 \cdot 4}$

Add $4$ and $0$ .

$x = \frac{- 2 \pm \sqrt{4}}{2 \cdot 4}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm \sqrt{2^{2}}}{2 \cdot 4}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 2 \pm 2}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{- 2 \pm 2}{8}$

Simplify $\frac{- 2 \pm 2}{8}$ .

$x = \frac{- 1 \pm 1}{4}$

Step 4

The final answer is the combination of both solutions.

$x = 0, - \frac{1}{2}$